1+2n(n-2)=-3(1-8n)

Solution for 1+2n(n-2)=-3(1-8n) equation:

1+2n(n-2)=-3(1-8n)

We move all terms to the left:

1+2n(n-2)-(-3(1-8n))=0

We add all the numbers together, and all the variables

2n(n-2)-(-3(-8n+1))+1=0

We multiply parentheses

2n^2-4n-(-3(-8n+1))+1=0


We calculate terms in parentheses: -(-3(-8n+1)), so:

-3(-8n+1)

We multiply parentheses

24n-3

Back to the equation:

-(24n-3)


We get rid of parentheses

2n^2-4n-24n+3+1=0

We add all the numbers together, and all the variables

2n^2-28n+4=0

a = 2; b = -28; c = +4;
Δ = b2-4ac
Δ = -282-4·2·4
Δ = 752
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{752}=\sqrt{16*47}=\sqrt{16}*\sqrt{47}=4\sqrt{47}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4\sqrt{47}}{2*2}=\frac{28-4\sqrt{47}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4\sqrt{47}}{2*2}=\frac{28+4\sqrt{47}}{4}
$